Question: Let $h(x)=x^4+4x^3-9$. What is the absolute minimum value of $h$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-64$ (Choice B) B $-36$ (Choice C) C $-72$ (Choice D) D $h$ has no minimum value
Solution: Let's first find the relative extremum points of $h$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $h$. The derivative of $h$ is $h'(x)=4x^2(x+3)$. $h'(x)=0$ for $x=0,-3$. $h'$ is defined for all real numbers. Therefore, our critical points are $x=0$ and $x=-3$. Our critical points divide the function's domain (which is all real numbers) into three intervals: $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $(-\infty, \llap{-}3)$ $( \llap{-}3,0)$ $(0,\infty)$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $(-\infty,-3)$ $x=-4$ $h'(-4)=-64<0$ $h$ is decreasing $\searrow$ $(-3,0)$ $x=-2$ $h'(-2)=16>0$ $h$ is increasing $\nearrow$ $(0,\infty)$ $x=1$ $h'(1)=16>0$ $h$ is increasing $\nearrow$ Now let's look at all the critical points: $x$ $h(x)$ Before After Verdict $-3$ $-36$ $\searrow$ $\nearrow$ Minimum $0$ $-9$ $\nearrow$ $\nearrow$ Not an extremum Let's imagine ourselves walking on the graph of $h$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going down until $(-3,-36)$, and then forever go up. This means that $\lim_{x\to-\infty}h(x)=\lim_{x\to +\infty}h(x)=+\infty$, which means $h$ has no maximum value. However, $h$ does reach an absolute minimum point at $(-3,-36)$, which means its absolute minimum value is $-36$. In conclusion, the absolute minimum value of $h$ is $-36$.